Input interpretation
3-bromo-2-(3'-chlorobenzyloxy)-5-methylphenylboronic acid | molar mass
Result
Find the molar mass, M, for 3-bromo-2-(3'-chlorobenzyloxy)-5-methylphenylboronic acid: M = sum _iN_im_i Plan: • Write the chemical formula and gather atomic masses from the periodic table. • Determine values for N_i and m_i using these items. • Finally, compute the mass. Write the chemical formula: C_14H_13BBrClO_3 Use the chemical formula, C_14H_13BBrClO_3, to count the number of atoms, N_i, for each element: | N_i Br (bromine) | 1 C (carbon) | 14 B (boron) | 1 O (oxygen) | 3 Cl (chlorine) | 1 H (hydrogen) | 13 Look up the atomic mass, m_i, in g·mol^(-1) for each element in the periodic table: | N_i | m_i/g·mol^(-1) Br (bromine) | 1 | 79.904 C (carbon) | 14 | 12.011 B (boron) | 1 | 10.81 O (oxygen) | 3 | 15.999 Cl (chlorine) | 1 | 35.45 H (hydrogen) | 13 | 1.008 Multiply N_i by m_i to compute the mass for each element. Then sum those values to compute the molar mass, M: Answer: | | | N_i | m_i/g·mol^(-1) | mass/g·mol^(-1) Br (bromine) | 1 | 79.904 | 1 × 79.904 = 79.904 C (carbon) | 14 | 12.011 | 14 × 12.011 = 168.154 B (boron) | 1 | 10.81 | 1 × 10.81 = 10.81 O (oxygen) | 3 | 15.999 | 3 × 15.999 = 47.997 Cl (chlorine) | 1 | 35.45 | 1 × 35.45 = 35.45 H (hydrogen) | 13 | 1.008 | 13 × 1.008 = 13.104 M = 79.904 g/mol + 168.154 g/mol + 10.81 g/mol + 47.997 g/mol + 35.45 g/mol + 13.104 g/mol = 355.42 g/mol
Unit conversion
0.3554 kg/mol (kilograms per mole)
Comparisons
≈ 0.49 × molar mass of fullerene ( ≈ 721 g/mol )
≈ 1.8 × molar mass of caffeine ( ≈ 194 g/mol )
≈ 6.1 × molar mass of sodium chloride ( ≈ 58 g/mol )
Corresponding quantities
Mass of a molecule m from m = M/N_A: | 5.9×10^-22 grams | 5.9×10^-25 kg (kilograms) | 355 u (unified atomic mass units) | 355 Da (daltons)
Relative molecular mass M_r from M_r = M_u/M: | 355